Simplify; express your answer in exponential form. Assume $q\neq 0, r\neq 0$. $\dfrac{{(q^{3})^{3}}}{{(q^{-5}r^{2})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{3}}$ to the exponent ${3}$ . Now ${3 \times 3 = 9}$ , so ${(q^{3})^{3} = q^{9}}$ In the denominator, we can use the distributive property of exponents. ${(q^{-5}r^{2})^{3} = (q^{-5})^{3}(r^{2})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{3})^{3}}}{{(q^{-5}r^{2})^{3}}} = \dfrac{{q^{9}}}{{q^{-15}r^{6}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{9}}}{{q^{-15}r^{6}}} = \dfrac{{q^{9}}}{{q^{-15}}} \cdot \dfrac{{1}}{{r^{6}}} = q^{{9} - {(-15)}} \cdot r^{- {6}} = q^{24}r^{-6}$.